Parallel Plate Capacitor Calculator
Use our online calculator to determine the capacitance of a parallel plate capacitor. Learn the formula, principles, and key variables like plate area and distance.
Parameters
Value for vacuum is 1. Select a material from the table below.
Capacitor Diagram
Result
Capacitance (C)
-- pF
Visualizer
The capacitance of a parallel plate capacitor is determined by the physical properties of its conductive plates and the dielectric material between them. This relationship is described by the following formula:
C = κ * ε₀ * (A / d)
- C is the capacitance, measured in Farads (F).
- κ (kappa) is the dielectric constant (or relative permittivity) of the material between the plates. It is a dimensionless quantity.
- ε₀ (epsilon-naught) is the permittivity of free space, a constant value of approximately 8.854 x 10-12 F/m.
- A is the overlapping surface area of the two plates, measured in square meters (m²).
- d is the separation distance between the plates, measured in meters (m).
Frequently Asked Questions
What is the primary function of a capacitor?
A capacitor's primary function is to store electrical energy in an electric field. It can charge up to store energy and then discharge to release that energy quickly, making it useful in applications like camera flashes, power supply smoothing, and signal filtering.
How does a dielectric material increase capacitance?
A dielectric material is an insulator. When placed between the capacitor plates, its molecules polarize, creating a small electric field that opposes the field from the plates. This reduces the overall electric field and voltage between the plates for a given amount of charge, allowing more charge (and thus more energy) to be stored. The factor by which it increases capacitance is its dielectric constant, κ.
What happens if the distance between the plates is doubled?
As seen in the formula C = κε₀A/d, capacitance (C) is inversely proportional to the distance (d). If you double the distance between the plates, the capacitance will be halved, assuming the area and dielectric remain constant.
